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Radius/interval of convergence

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\Sigma[/tex] n!(2x-1)[tex]^{}n[/tex]

    from n=1 to infinity

    2. Relevant equations

    -ratio test

    3. The attempt at a solution

    lim n-> infinity | (n+1)!(2x-1)^(n+1) / n!(2x-1)^n |

    lim n-> infinity | (n+1)(2x-1) |

    |2x - 1| lim n-> infinity | (n+1) |

    -1 < 2x -1 < 1

    0 < 2x < 2
    0 < x < 1

    I know at this point I've done something pretty wrong. as my interval and radius doesnt match up with the back of the book. could use a push in the right direction, thanks.
  2. jcsd
  3. Apr 24, 2010 #2


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    What happened to the lim n->infinity |n+1|? Did you just drop it?
  4. Apr 24, 2010 #3
    oh, im not too sure what to do with it. i understand that the limit goes to infinity, UNLESS x = 1/2 in which case it goes to 0. how do i proceed knowing that?

    nm, i should just stop there ;)

    1/2 is the only value of x for which this will converge, so the interval of convergence is {1/2}. the radius is 0.
    Last edited: Apr 24, 2010
  5. Apr 24, 2010 #4


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    I couldn't have said it better myself.
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